45406 + Interview Questions in Physics in Class 12

1.	Which of the following atoms behave as polar dielectric ?
Answer» `H_(2)` and `H_(2)O` `O_(2)` and `H_(2)O` `CO_(2)` and `H_(2)O` HCI and `H_(2)O` Answer :A::B::C::D

Discussion

2.	Beats result from two vibratory motions: s_(1)=cos4999pit and s_(2)=cos5001pit Find the period of beat and the “conventional" period of the almost sinusoidal vibrations.
Answer» ANSWER :0.5 s; 0.4 MS.

Discussion

3.	The following series L-C-R circuit, when driven by an e.m.f. source of angular frequency 70 kilo -radians per second, the circuit effectively behaves like
Answer» PURELY RESISTIVE CIRCUIT series R -L circuit series R - C circuit series L- C circuit with R = O Answer :C

Discussion

4.	In Young's double-slit experiment, the fringe pattern shifts by a distance x_(0) when a mica sheet, 1.964 mu m thich and of refractive index 1.6, covers one of the slits. If the mica sheet is removed and the slite-to-screen distance doubled, the new fringe width is equal to x_(0). Find the wavelength of light used.
Answer» Solution :Data: `n_(m) = 1.6, B=1.964, mu m= 1.964 xx 10^(-6)m, D_(2) = 2D_(1), X_(2) = x_(0)` The fringe shift with the mica sheet, `x_(0) = D_(1)/d (n_(m)-1)b` Subsequent to the removal of the mica sheet and doubling the slits-to-screen DISTANCE, the new fringe width is `X_(2) = (lambdaD_(2))/d = (lambda(2D_(1)))/(d)` Since, `X_(2) = x_(0)`, `(2lambdaD_(1))/(d) = D_(1)/d(n_(m)-1)b` `THEREFORE` The wavelength of the LIGHT used, `lambda= (n_(m)-1)/(2).b-(1.6-1)/2.(1.964 xx 10^(-6))` `=0.3 xx 1.964 xx 10^(-6) = 5.892 xx 10^(-7)m = 5892 Å`

Discussion

5.	The wavelength less than …… Å almost all the electromagnetic waves are absorb in ozone layer.
Answer» `4000 Å` `3000 Å` `5000 Å` `6000 Å` Answer :B

Discussion

6.	For an equilateral prism made of a material of Refractive index sqrt2, find the angle of minimum deviation for a ray of monochromatic light.
Answer» SOLUTION :`n=SQRT2` and `A=60^@` `n=[SIN(A+D)/2]/(sinA/2) sqrt2=[sin(60+D)/2]/SIN30^@` `sqrt2/2=sin(60+D)/2 1/sqrt2=sin((60+D)/2)` `(60+D)/2=45, D=30^@`

Discussion

7.	Suppose drift velocity over the entire cross section of a wire isV (r) = V_(0) [ 1 - (r)/(R) ] . What is the current density at the surface of the wire ?
Answer» zero `V_(0)`R ne`V_(0)` ne`(r)/(R)` Solution :zero `V_(r) = V_(0) [ 1 - (r)/(R) ] ` on the surface , taking r = R. `therefore V_((r))= V_(0) [ 1- 1 ] ` `therefore V_((r)) = V_(0) [0] =0 `

Discussion

8.	First overtone frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. Further nth harmonic of closed organ pipe is also equal to the mth harmonic of open pipe, where n and m arc:
Answer» 5,4 7,5 9,6 7,3 ANSWER :C

Discussion

9.	A particle is moving in such a way that its velocity versus time graph is a parabola (t=kv^(2),k is constant) as shown in the figure. Choose the correct option(s)
Answer» ACCELERATION of the particle is `sqrt(3)m//s^(2)` throughout the journey average acceleration of particle is `1m//s^(2)` for the INTERVAL of first 5 sec average acceleration of particle is LESS than `sqrt(3)m//s^(2)` for the interval of first 5 sec average acceleration of particle is less than `1m//s^(2)` for the interval of 5 sec to 10 sec Solution :AVG. acceleration `=("change in velocity")/("TIME interval")`

Discussion

10.	What is de-Broglie wave?
Answer» SOLUTION :The wave associated with MATERIAL particle is CALLED de-Broglie wave

Discussion

11.	Figure shows a charge array known as an "electric quadrupole". For a point on the axis of the quadruple, the dependence of potential on r(r/agt gt1) is 1/(r^(n)). Find n?
Answer» SOLUTION :3

Discussion

12.	The p.d. between A and B is balanced at 2.036 m length of the potentiometer wire. When the balancing length between A and C is 0.246 m, then between C and B will be. (##MOT_CON_NEET_PHY_C22_E01_033_Q01.png" width="80%">
Answer» `1.79` `2.79` `0.79` `2.49` ANSWER :A

Discussion

13.	Find the effective resistance of series connection of parallel resistors.
Answer» Solution :(i) Suppose n CELLS, each of emf `xi` volts and internal resistance r ohms are connected in series with an external resistance R. as shown in Figure. (ii) The total emf of the battery `n xi` The total resistance in the circuit `nr+R` By Ohm's law, the current in the circuit is `I=("total emf")/("total resistance") =(n xi)/(nr+R)` Case (a) If `r lt lt R` then, `I=(n xi)/(R)~~ nI_(1)` where `I_(1)` is the current due to a single cell `(I_(1)=(xi)/(R))` (iii) Thus, if r is negligible when compared to R the current supplid by the battery is n times that supplied by a single cell. Case (b) a single cell. Case (b) If `r gt gt R, I=(n xi)/(nr)~~(xi)/(r)` (iv) It is the current due to a single cell. That is, current due to the whole batteryis the same as that due to a single cell and hence there is no advantage in connecting several cell. (v) Thus series connection of the cell is ADVANTAGEOUS only when the effective internal resistance of the cells is negaligibly smallcompared with R. Cells in PARALLEL (i) In parallel connection all the positive terminnals of the cells are connected toone point and all the negative terminals to a second point. These two points from the positive and negative terminals of the battery. (ii) Let n cells be connected in parallel between the points A and B and a resistance R is connected between the point a and B as shown in figure. Let `xi` be the emf and r the internal resistance of each cell. (iii) The equivalent internal resistance of the battey is `(1)/(r_(eq))=(1)/(r)+(1)/(r)+....(1)/(r)` (NTERMS) =`(n)/(r)` So `(I)/(r_(eq))=(r)/(n)` and the total resistance in the circuit `R+(r)/(n)`. The total emf is the potential difference between the points A and B, which is equal to `xi` the current in the circuite is given by `I=(xi)/((r)/(n)+R)` `I=(n xi)/(r+nR)` Case (a) If `r gt gt R, I=(n xi)/(r)=nI_(1)` (iv) Where `I_(1)` is the current due to a single cell and is equal to `(xi)/(r)` when R is negligible. Thus, the current through the externla resistance due to the whole battery is n times the current due to a single cell. Case (b) If `r lt lt R, I(xi)/(R)` (v)The above equation implies that current due to the whole battery is the same as that due to a single cell. Hence it is advantageous to connecte cells in parallel when the external resistance is very small compared to the internal of the cells.

Discussion

14.	Moment of inertia of a body about two perpendicular axes X and Y in the plane of lamina are 20" kg m"^2and 25 "kg m"^2respectively. Its moment of inertia about an axis perpendicular to the plane of the lamina and passing through the point of intersection of X and Y axes is
Answer» `5kgm^(2)` `45kgm^(2)` `12-5kgm^(2)` `500KGM^(2)` Solution :PERPENDICULAR AXIS tehorem `I_(Z)=I_(x)+I_(y)=20+25=45kgm^(2)`

Discussion

15.	गंगा और जमुना का संगम कहां पर होता है?
Answer» हरिद्वार इलाहाबाद वाराणसी देवप्रयाग Answer :B

Discussion

16.	नमामि गंगे परियोजना किस नदी से संबंधित है?
Answer» गंगा नदी नर्मदा कावेरी नदी ब्रह्मपुत्र नदी Answer :A

Discussion

17.	Considerthe chargesq,q and -q placed at the verticesof an equilateraltrianglewhat is the force on each charge
Answer» Solution :The forces actingon charge q at a due to charges q at B and -q at c are `F_(12)`along BA and `F_(13)` along AC respectivelyq at a is given by the force of attaraction or repulson for each pari of charges has thesame magnitude `F=(q^(2))/(4piepsilon_(0)L^(2))` The totalforce `f_(2)` on chargeq at b is thus `f_(2)` =f `vecr_(2)`where `vecr_(2)` is a unit vectoralong AC `F_(1)+F_(2)+F_(3)=0` The result is not at all surprisingit follows straightfrom the fact that colulomb law is CONSISTENT with newton thrid law the proof is left to you as an exercise

Discussion

18.	The activity of a certain radioactive sample is plotted against time in graph. If the initial slope of the curve is m, then find the slope of the curve at point P ?
Answer» <P> Solution :Activity `A=A_0 e^(-lambdat)` Slope of the graph is `(dA)/(dt)=-lambdaA_0e^(-lambdat)` Initially `t=0 RARR (dA)/(dt)=-lambdaA_0` (=m) given Slope of POINT P is , `(dA)/(dt)=-lambdaA_0 e^(-lambdatau)` `=m/3 [ because lambdatau=1]`

Discussion

19.	Of the following, which is the best description of the second law of thermodynamics?
Answer» The total energy of the UNIVERSE is a constant. The efficiency of a heat engine can never be greater than 50 percent. The amount of heat required to vaporise a liquid is greater than the amount of heat required to melt a solid of the same substance. The entropy of the universe is always increasing. Solution :Choice (A) is the first law of thermodynamics. CHOICES (C) and (D) are true, but they are not EQUIVALENT to the second law of thermodynamics. Choice (B) is FALSE, but if it read, "The efficiency of a heat engine can never be equal to 100 percent", then it would be equivalent to the second law.

Discussion

20.	Whether alpha or beta rays are more penetrating ?
Answer» SOLUTION :`BETA`-RAYS

Discussion

21.	The law of conservation of momentum applies to a system of colliding objects only if
Answer» there is no change in KINETIC energy of the system. the COEFFICIENT of restitution is one. the coefficient of restitution is zero the NET EXTERNAL IMPULSE is zero. Answer :D

Discussion

22.	The AM wave contains three frequencies , viz :
Answer» `(f_(C))/(2) , (f_(c) + f_(s))/(2) , (f_(c) - f_(s))/(2)` `2 f_(c) , 2 (f_(c) + f_(s)) , 2 (f_(c) - f_(s))` `f_(c) , (f_(c) + f_(s)) , (f_(c) - f_(s))` `f_(c) , f_(c) ,f_(c)` ANSWER :C

Discussion

23.	निम्न में से तत्व का अणु किसमें होगा
Answer» कार्बन डाई ऑक्साइड सल्फर डाइऑक्साइड `N_2` `NH_3` ANSWER :C

Discussion

24.	Kirchhoff's junction rule is based on the law of conservation of electrical energy.
Answer» SOLUTION : FALSE - Kirchhoff.s JUNCTION rule is based on the law of conservation of electric charge.

Discussion

25.	परमाणु का द्रव्यमान किसके सापेक्ष मापा जाता है
Answer» C-14 C-12 O-16 N-14 Answer :B

Discussion

26.	The capacitors in figure are initially uncharged and are connected as in the diagram with switch S open. The applied potential difference is V_(ab)= 360 V. How much charge flowed through the switch when it was closed
Answer» `140 MUC` from C to D `240 MU` from C to D `540 mu` from C to D `120 muC` from C to D Answer :C

Discussion

27.	The capacitors in figure are initially uncharged and are connected as in the diagram with switch S open. The applied potential difference is V_(ab)= 360 V. What is the potential difference across each capacitor after switch S is closed
Answer» 100 V 180V 200 V 300 V Answer :B

Discussion

28.	The capacitors in figure are initially uncharged and are connected as in the diagram with switch S open. The applied potential difference is V_(ab)= 360 V. Plates of a capacitor have charges 2C_(epsilon) and C_(epsilon)initially. Now the switch S is closed, Which of the following statements is true? Assume that the battery, resistor and wires do not have any capacitance.
Answer» There is no change in charge at outer surfaces of the capacitor. There is no change in charge inner surfacesof the capacitor. Charge flowing through battery is ZERO. Charge flowing through the battery is`(C_(EPSILON))/2` ANSWER :A::D

Discussion

29.	The capacitors in figure are initially uncharged and are connected as in the diagram with switch S open. The applied potential difference is V_(ab)= 360 V. In the given circuit diagram, find the charge which will flow through direction 1 and 2 when switch S is closed.
Answer» `q_(1)=EC_(2),q_(2)=-(EC_(1)C_(2))/((C_(1)+C_(2)))` `q_(1)=EC_(1),q_(2)=-(EC_(2)C_(1))/((C_(1)-C_(2)))` `q_(1)=EC_(1),q_(2)=-(EC_(2))/((C_(1)+C_(2)))` `q_(1)=EC_(1),q_(2)=-(EC_(2))/((C_(1)+C_(2)))` Answer :A

Discussion

30.	Electromagnetic waves are transverse in nature is evident by
Answer» POLARIZATION Interference Reflection Diffraction. Solution :Polarization is SHOWN by only TRANSVERSE WAVES.

Discussion

31.	Why does a driver use a convex mirror as a rear view mirror?
Answer» SOLUTION :The IMAGE formed by convex mirror is always virtual for all the POSITIONS of the object in front of it. The image is always diminished for all positons of the object in front of it. So a driver can see all the images in the mirror irrespective of the position of OBJECTS.

Discussion

32.	Potential difference across the terminals of a cell becomes zero when the terminals are connectedby a metallic wire.
Answer» Solution :TRUE - As resistance of METALLIC WIRE is zero, CURRENT in the circuit is I =`epsi/r`and hence `V = epsi- Ir = epsi-epsi = 0`.

Discussion

33.	The capacitors in figure are initially uncharged and are connected as in the diagram with switch S open. The applied potential difference is V_(ab)= 360 V. What is the potential difference V_(cd) ?
Answer» 120 V 100V 150 V 115 V Answer :A

Discussion

34.	A body is released from a great height and falls freely under gravity vertically towards eurth. Another body is released from the same height exactly one second later from the same point. Separation between the two after 2 second of release of the second body is :
Answer» 24.5 m 14.6 m 9.8 m 4.9 m Solution :For first body , `h_1=(1)/(2)G t^(2)` For SECOND body `h_2=(1)/(2)g(t-1)^(2)` For SEPARATION `h_1-h_2=(1)/(2)[t^(2)=(t-1)^(2)]` HENCE t=(2+1)=3s `:. H_1-h_2=(9.8)/(2)[9-4]=24.5m`

Discussion

35.	In a Young's double slit experiment, the intensity at a point where the path difference is (lamda)/6 (lamda- wavelength of the light) is 1. IfI_(0) denotes the maximum intensity, then I/(I_(0)) is equal to
Answer» `1/2` `(SQRT(3))/2` `1/(sqrt(2))` `3/4` ANSWER :D

Discussion

36.	The nucleus ""_(92)^(235)Y , initially at rest, decays into ""_(90)^(213)X by emitting an alpha-particle ""_(92)^(235)Y to ""_(90)^(231)X + energy, The binding energies per nucleon of the parent nucleus, the daughter nucleus and alpha-particle are 7.8 MeV, 7.835 MeV and 7.07 MeV, respectively. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, find the speed of the emitted alpha-particle, ["Mass of particle "=6.68 xx 10^(-27) kg]
Answer» Solution :In the nuclear reaction `""_(92)^(235)Y to ""_(90)^(231)X+""_(2)^(4)He`+ energy, the energy released will be E=Total Binding energy of products-Total Binding energy of PARENT NUCLEUS `=(231 xx 7.835+4 xx 7.07-235 xx 7.8) MeV` =1809.885+28.280-183.0=5.165MeV `=5.165 xx 1.6 xx 10^(-13)J` This energy appearns as the KINETIC energy of alpha particle of mass `m=6.68 xx 10^(-27)KG`. Iv v be the speed of alpha particle, then `E=1/2 mv^(2)` `rArr v=sqrt((2E)/(m))= [(2 xx 5.165 xx 1.6 xx 10^(-13))/(6.68 xx 10^(-27))]^(1/2) =4.97 xx 10^(7)ms^(-1)`

Discussion

37.	In the above question what is the distance covered bythe particle in 5th second of its motion ?
Answer» 4 m 8 m 5 m 11 m Solution :`x=(t-2)^(2)=t^(2)+4-4` `(dx)/(dt)=v=2t-4` `:.` Velocity u at t=0is u=-4 `ms^(-1)` ALSO `(dv)/(dt)=a=2 ms^(-2)` Thus the body has uniform acceleration. Here `S_(n)=u+(a)/(2)(2n-1)` `=-4+2//2(2xx5-1)` `= -4+9=5 m`

Discussion

38.	The branch of physics which deals with static electric charges or charges at rest is ............. .
Answer» ANSWER :electronstatics

Discussion

39.	In long distance communication, the similar potential wave produced from a carrier wave is V = V_(0) sin Omegat. On the other side, potential wave corresponding to data signal is , v = v_(0) sin omega t, "" v_(0) lt lt Omega. Superimposing data signal on carrier wave, generally two types of modulated waves are produced, they are: (a) Amplitude Modulated (AM) wave: V_("AM") = V_(0) (1 + beta_(1) sin omega t ) sin Omega t where, beta_(1) = k_(1)(v_(0))/(V_(0)) = modulation index ""(k_(1) " " constant). Frequency Modulated (FM) wave: V_("FM") = V_(0) sin(Omegat - beta_(2) cos omega t ) where, beta_(2) = k_(2)(v_(0))/(omega) = modulation index "" (k_(2) = constant). Mathematical analysis of two modulated waves shows that, for AM wave, apart from Omega frequency, there are two sine wave components of frequencies Omega - omega and Omega + omega. As a result, the bandwidth of the waves becomes (Omega + omega)-(Omega-omega) = 2omega. On the other hand, in FM wave, the number of sine wave components are infinite and their frequencies areOmegan pm, Omega pm 2omega,... . Basides that, if the dissipated power for a transmitting antenna to transmit a carrier wave be P_(c ) and dissipated power for transmitting AM and FM wave be P_(AM)and P_(FM) then, P_(AM) = P_(c )(1 + (beta^(2))/(2)), P_(FM) = P_(c ) What would be the bandwidth of an amplitude modulated wave if there is a mixing of frequencies 1000 Hz to 1500 Hz in the data signal?
Answer» 500 Hz 1500 Hz 2000 Hz 3000 Hz Solution :N/A

Discussion

40.	In long distance communication, the similar potential wave produced from a carrier wave is V = V_(0) sin Omegat. On the other side, potential wave corresponding to data signal is , v = v_(0) sin omega t, "" v_(0) lt lt Omega. Superimposing data signal on carrier wave, generally two types of modulated waves are produced, they are: (a) Amplitude Modulated (AM) wave: V_("AM") = V_(0) (1 + beta_(1) sin omega t ) sin Omega t where, beta_(1) = k_(1)(v_(0))/(V_(0)) = modulation index ""(k_(1) " " constant). Frequency Modulated (FM) wave: V_("FM") = V_(0) sin(Omegat - beta_(2) cos omega t ) where, beta_(2) = k_(2)(v_(0))/(omega) = modulation index "" (k_(2) = constant). Mathematical analysis of two modulated waves shows that, for AM wave, apart from Omega frequency, there are two sine wave components of frequencies Omega - omega and Omega + omega. As a result, the bandwidth of the waves becomes (Omega + omega)-(Omega-omega) = 2omega. On the other hand, in FM wave, the number of sine wave components are infinite and their frequencies areOmegan pm, Omega pm 2omega,... . Basides that, if the dissipated power for a transmitting antenna to transmit a carrier wave be P_(c ) and dissipated power for transmitting AM and FM wave be P_(AM)and P_(FM) then, P_(AM) = P_(c )(1 + (beta^(2))/(2)), P_(FM) = P_(c ) The unit constant K_(2)
Answer» 1 `V CDOT s^(-1)` `V^(-1) cdot s` `V^(-1) cdot s^(-1)` SOLUTION :N/A

Discussion

41.	In long distance communication, the similar potential wave produced from a carrier wave is V = V_(0) sin Omegat. On the other side, potential wave corresponding to data signal is , v = v_(0) sin omega t, "" v_(0) lt lt Omega. Superimposing data signal on carrier wave, generally two types of modulated waves are produced, they are: (a) Amplitude Modulated (AM) wave: V_("AM") = V_(0) (1 + beta_(1) sin omega t ) sin Omega t where, beta_(1) = k_(1)(v_(0))/(V_(0)) = modulation index ""(k_(1) " " constant). Frequency Modulated (FM) wave: V_("FM") = V_(0) sin(Omegat - beta_(2) cos omega t ) where, beta_(2) = k_(2)(v_(0))/(omega) = modulation index "" (k_(2) = constant). Mathematical analysis of two modulated waves shows that, for AM wave, apart from Omega frequency, there are two sine wave components of frequencies Omega - omega and Omega + omega. As a result, the bandwidth of the waves becomes (Omega + omega)-(Omega-omega) = 2omega. On the other hand, in FM wave, the number of sine wave components are infinite and their frequencies areOmegan pm, Omega pm 2omega,... . Basides that, if the dissipated power for a transmitting antenna to transmit a carrier wave be P_(c ) and dissipated power for transmitting AM and FM wave be P_(AM)and P_(FM) then, P_(AM) = P_(c )(1 + (beta^(2))/(2)), P_(FM) = P_(c ) In converting a carrier wave into an Amplitude Modulated (AM), modulation index was 0.05. Then, what would be the percentage increases in dissipated power at transmitting antenna?
Answer» 2.5 1.25 0.25 0.125 Solution :N/A

Discussion

42.	In long distance communication, the similar potential wave produced from a carrier wave is V = V_(0) sin Omegat. On the other side, potential wave corresponding to data signal is , v = v_(0) sin omega t, "" v_(0) lt lt Omega. Superimposing data signal on carrier wave, generally two types of modulated waves are produced, they are: (a) Amplitude Modulated (AM) wave: V_("AM") = V_(0) (1 + beta_(1) sin omega t ) sin Omega t where, beta_(1) = k_(1)(v_(0))/(V_(0)) = modulation index ""(k_(1) " " constant). Frequency Modulated (FM) wave: V_("FM") = V_(0) sin(Omegat - beta_(2) cos omega t ) where, beta_(2) = k_(2)(v_(0))/(omega) = modulation index "" (k_(2) = constant). Mathematical analysis of two modulated waves shows that, for AM wave, apart from Omega frequency, there are two sine wave components of frequencies Omega - omega and Omega + omega. As a result, the bandwidth of the waves becomes (Omega + omega)-(Omega-omega) = 2omega. On the other hand, in FM wave, the number of sine wave components are infinite and their frequencies areOmegan pm, Omega pm 2omega,... . Basides that, if the dissipated power for a transmitting antenna to transmit a carrier wave be P_(c ) and dissipated power for transmitting AM and FM wave be P_(AM)and P_(FM) then, P_(AM) = P_(c )(1 + (beta^(2))/(2)), P_(FM) = P_(c ) What would be the width of sideband on either side of the amplitude modulated (AM) wave, if there is a mixing of frequencies of 1000 Hz to 1500 Hz in the data signal?
Answer» 500 Hz 1000 Hz 1500 Hz 2000 Hz Solution :N/A

Discussion

43.	The temperature co.efficient of resistance for a wire is 0.00125^@C.At 300K ,it's resistance is 1 ohm.The temperature at which the resistance becomes 2 ohm is :
Answer» 1154 K 1100 K 1400 K 1127 K Answer :D

Discussion

44.	Two sinusoidal waves with the same amplitude of 6.00 mm and the same wavelength travel together along a string that is stretched along an x axis. Their resultant wave is shown twice in Fig. 16 39. as velley travels in the negative direction of the x axis by distance d= 56.0 cm in 8.0 ms. The lick matks along the axis are separated by 10cm, and height H is 8.0mm. Let the equation for one wive be of the form y(x,t) =y_m sin (kx pm omegat+phi_(1))," where "phi_(1)=0and you must choose the correct sign in front of omega. For the equation for the other wave, what are (a) y_(m), (b) k, (c) omega, (d) phi_(2), and (e) the sign in front of omega?
Answer» ANSWER :(a) 6.00 mm; (B) 16rad/m; (C) `1.1 XX 10^(3)" rad/s"` (d) 2.5 rad; (e) PLUS sign

Discussion

45.	The main purpose for the classification of organisms is to
Answer» STUDY geography locate PLANTS and animals establish RELATIONSHIPS AMONGST organisms study evolution. Answer :C

Discussion

46.	For a gas sample with N_(0) number of molecules, function N(v) is given by : N(v)=(dN)/(dv)=((3N_(0))/(v_(0)^(3)))v^(2)" for "0ltvltv_(0)andN(v)=0" for "vgtv_(0), where dN is number of molecules in speed range v to v+dv. Find the rms speed of the molecules.
Answer» `sqrt((5)/(3))v_(0)` `sqrt((3)/(5))v_(0)` `sqrt((5)/(3v_(0)))` `sqrt((3)/(5v_(0)))` Solution :`v_(RMS)^(2)=ltv^(2)GT =(v_(1)^(2)+v_(2)^(2)+v_(3)^(2)+....)/(N)=(intv^(2)dN)/(intdN)` Here `(dN)/(dv)=N(v)`, So, `v_(rms)^(2)=(1)/(N)underset(0)OVERSET(oo)intN(v)v^(2)dv=(1)/(N)underset(0)overset(v_(0))int((3N)/(v_(0)^(3)).v^(2))v^(2)dv=(3)/(5)v_(0)^(2)` `implies v_(rms)=sqrt((3)/(5))v_(0)`

Discussion

47.	What is the unit of capacitance?
Answer» Solution :The UNIT of CAPACITANCE is farad. It is ABBREVIATED by F.

Discussion

48.	a. Where on the earth's surface is the value of vertical component of the earth's magnetic field zero ? b. The horizontal component of the earth's magnetic field at a given place is 0.4xx10^(-4) Wb//m^2 and angle of dip is 30^@ Calculate the value of (i) vertical component (ii) the total intensity of the earth's magnetic field.
Answer» Solution :a. At the equator , the value of VERTICAL component of earth.s MAGNETIC field is zero. b. i. `Z_E=B_E SIN I = 0.4 xx10^(-4) xx tan 30^@ = 0.23 xx10^(-4) Wbm^(-2)` ii. TOTAL intensity of earth.s magnetic field, `B_E=H_E/(COSI)=(0.4xx10^(-4))/(cos30^@)=0.46xx10^(-4) Wbm^(-2)`

Discussion

49.	In young's experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 2 micron is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima now is the same as the observed fringe shit upon the introduction of the mica sheet. The wavelength of light is
Answer» `5762 A^(@)` `5825A^(@)` `6000A^(@)` `6500A^(@)` SOLUTION :6000A

Discussion

50.	The mathematical expression for Lenz's law is given by
Answer» SOLUTION :E = `[-Ndphi/dt]`

Discussion

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For an equilateral prism made of a material of Refractive index sqrt2, find the angle of minimum deviation for a ray of monochromatic light.

Suppose drift velocity over the entire cross section of a wire isV (r) = V_(0) [ 1 - (r)/(R) ] . What is the current density at the surface of the wire ?

First overtone frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. Further nth harmonic of closed organ pipe is also equal to the mth harmonic of open pipe, where n and m arc:

A particle is moving in such a way that its velocity versus time graph is a parabola (t=kv^(2),k is constant) as shown in the figure. Choose the correct option(s)

What is de-Broglie wave?

Figure shows a charge array known as an "electric quadrupole". For a point on the axis of the quadruple, the dependence of potential on r(r/agt gt1) is 1/(r^(n)). Find n?

The p.d. between A and B is balanced at 2.036 m length of the potentiometer wire. When the balancing length between A and C is 0.246 m, then between C and B will be. (##MOT_CON_NEET_PHY_C22_E01_033_Q01.png" width="80%">

Find the effective resistance of series connection of parallel resistors.

Moment of inertia of a body about two perpendicular axes X and Y in the plane of lamina are 20" kg m"^2and 25 "kg m"^2respectively. Its moment of inertia about an axis perpendicular to the plane of the lamina and passing through the point of intersection of X and Y axes is

गंगा और जमुना का संगम कहां पर होता है?

नमामि गंगे परियोजना किस नदी से संबंधित है?

Considerthe chargesq,q and -q placed at the verticesof an equilateraltrianglewhat is the force on each charge

The activity of a certain radioactive sample is plotted against time in graph. If the initial slope of the curve is m, then find the slope of the curve at point P ?

Of the following, which is the best description of the second law of thermodynamics?

Whether alpha or beta rays are more penetrating ?

The law of conservation of momentum applies to a system of colliding objects only if

The AM wave contains three frequencies , viz :

निम्न में से तत्व का अणु किसमें होगा

Kirchhoff's junction rule is based on the law of conservation of electrical energy.

परमाणु का द्रव्यमान किसके सापेक्ष मापा जाता है

The capacitors in figure are initially uncharged and are connected as in the diagram with switch S open. The applied potential difference is V_(ab)= 360 V. How much charge flowed through the switch when it was closed

The capacitors in figure are initially uncharged and are connected as in the diagram with switch S open. The applied potential difference is V_(ab)= 360 V. What is the potential difference across each capacitor after switch S is closed

The capacitors in figure are initially uncharged and are connected as in the diagram with switch S open. The applied potential difference is V_(ab)= 360 V. In the given circuit diagram, find the charge which will flow through direction 1 and 2 when switch S is closed.

Electromagnetic waves are transverse in nature is﻿ evident by﻿

Why does a driver use a convex mirror as a rear view mirror?

Potential difference across the terminals of a cell becomes zero when the terminals are connectedby a metallic wire.

The capacitors in figure are initially uncharged and are connected as in the diagram with switch S open. The applied potential difference is V_(ab)= 360 V. What is the potential difference V_(cd) ?

A body is released from a great height and falls freely under gravity vertically towards eurth. Another body is released from the same height exactly one second later from the same point. Separation between the two after 2 second of release of the second body is :

In a Young's double slit experiment, the intensity at a point where the path difference is (lamda)/6 (lamda- wavelength of the light) is 1. IfI_(0) denotes the maximum intensity, then I/(I_(0)) is equal to

In the above question what is the distance covered bythe particle in 5th second of its motion ?

The branch of physics which deals with static electric charges or charges at rest is ............. .

The temperature co.efficient of resistance for a wire is 0.00125^@C.At 300K ,it's resistance is 1 ohm.The temperature at which the resistance becomes 2 ohm is :

The main purpose for the classification of organisms is to

For a gas sample with N_(0) number of molecules, function N(v) is given by : N(v)=(dN)/(dv)=((3N_(0))/(v_(0)^(3)))v^(2)" for "0ltvltv_(0)andN(v)=0" for "vgtv_(0), where dN is number of molecules in speed range v to v+dv. Find the rms speed of the molecules.

What is the unit of capacitance?

The mathematical expression for Lenz's law is given by

Electromagnetic waves are transverse in nature is evident by